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jasenka [17]
3 years ago
8

Median and range of 11,17,19,6,17

Mathematics
1 answer:
Alexxandr [17]3 years ago
6 0
6,11,17,17,19
median is 17
11+17+19+6+17=70
70/5=14
mean is 14
19-6=13
13 is range
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Initially, there are 40 grams of A and 50 grams of B, and for each gram of B, 2 grams of A is used. It is observed that 15 grams
hram777 [196]

Answer:

X(16)=25.71grams

Step-by-step explanation:

let X(t) denote grams of C formed in  t mins.

For X grams of C we have:

\frac{2}{3}Xg of A and \frac{1}{3}Xg of B

Amounts of A,B remaining at any given time is expressed as:

40-\frac{2}{3}Xg of A and  50-\frac{1}{3}Xg  of B

Rate at which C is formed satisfies:

\frac{dX}{dt} \infty(40-\frac{2}{3}X)(50-\frac{1}{3}X)->\frac{dX}{dt}=k(90-X)\\\therefore \frac{dX}{(90-X)^2}=kdt->\int{\frac{dX}{(90-X)^2}} \, =\int {k} \, dt  \\\therefore \frac{1}{90-X}=kt+c->90-X=\frac{1}{kt+c}\\\\X(t)=90-\frac{1}{kt+c}

Apply the initial condition,X(0)=0 ,to the expression above

0=90-\frac{1}{c} \ \ ->c=\frac{1}{90}\\\therefore\\X(t)=90-\frac{1}{kt+\frac{1}{90}} \ \ ->X(t)=90-\frac{90}{90kt+c}

Now at X(8)=15:

15=90-\frac{90}{90\times 8k+1}  \ ->75=\frac{90}{720k+1}\\k=0.0002778

Substitute  in X(t) to get

X(t)=90-\frac{90}{0.0002778t\times 90+1}\\X(t)=90-\frac{90}{0.25t+1}\\But \ t=16\\\therefore X(t)=90-\frac{90}{0.025\times16+1}\\X(t)=25.71

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3 years ago
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Help pls i am being timed on a quiz
Dahasolnce [82]

In analyzing the given diagram which is a combination of two right angle triangle, the value of sin∠L is 5/13.

<h3>What is the value of sin∠L?</h3>

Given that angle ONM is a right angled triangle, we can find length of OM using the Pythagorean theorem.

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Next, angle OML is also a right angled triangle, we can get the hypotenuse LO.

c = √( a² + b² )

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LO = √( 5² + 12² )

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Now, to get sin∠L, we use the mnemonic: SOHCAHTOA

sinθ = opposite / hypotenuse

sin∠L = OM / LO

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Therefore, in analyzing the given diagram which is a combination of two right angle triangle, the value of sin∠L is 5/13.

Learn more about Pythagorean theorem here: brainly.com/question/343682

#SPJ1

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1 year ago
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